# Open coverings of $\mathbb{R}$

If at each rational number we take an open ball of constant size (that is the radius of the ball is the same for all rational numbers) then it seems that all these open balls cover $\mathbb{R}$ i.e, $\cup_{i=1}^\infty B(x_i,r) \supset \mathbb{R}$ (where $r > 0$ and $x_i$ are rationals).

If is the radius is not constant but arbitrary, then do the resulting open balls still cover $\mathbb{R}$ i.e., is it true that $\cup_{i=1}^\infty B(x_i,r_i) \supset \mathbb{R}$ (where $r_i > 0$)?

In general, is the above true for any separable metric space?

Thanks, Phanindra

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It's not true if $\sum_i r_i<\infty$ by a measure theory argument. –  Davide Giraudo May 12 '12 at 14:41
If $\{x_n\}_{n\in\mathbb{N}}$ is a dense subset of a separable metric space $X$ and $\{r_n\}_{n\in\mathbb{N}}$ is a sequence of positive real numbers, then for each $x\in X$ and for any open ball $B$ centered in $x$ you can find $x_n$ in $B$, but this doesnt mean that necessary $x\in B(x_n;r_n)$. –  Yuki May 12 '12 at 14:58

No. For instance if you enumerate the rationals $r_1,r_2,\ldots$ and choose the diameter of the $i^{\rm th}$ ball to be $\le 2^{-(i+1)}$, then the measure of the union of the balls will not exceed $1/2$. The complement of the union must then be non-empty.
Even easier, fix an irrational number $p$. Then for each rational number $q$, you can find an open ball centered on $q$ that excludes $p$. The union of these balls will exclude $p$ as well.
Or, you could consider a cover comprised of sets of the form $(n\pi,(n+1)\pi)$.